I have lost all notes for this and can't seem to work it out, although i'm sure it is very simple I apologise in advance!

A manufacturer has the following short-run total cost function:

$$TC = 100 + 25Q – 5Q^2 + Q^3$$

Derive the average variable cost (AVC) function and show that, when AVC is a minimum, marginal cost (MC) is equal to AVC.

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    $\begingroup$ You don't need notes for this: Total Variable Cost is Total Cost minus Fixed Cost (and what is Fixed Cost?). Average Variable Cost is Variable Cost divided by the function's argument. Then, "minimum" and "marginal" both somehow relate to the mathematical concept of "derivative". $\endgroup$ Dec 2, 2014 at 14:52
  • $\begingroup$ I'm not given anything other than the short-run total cost function... $\endgroup$ Dec 2, 2014 at 14:59
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    $\begingroup$ Which is all you need. Have you never encountered a function minimization problem before? What strange kind of Microeconomics (or Business microeconomics) course are you attending? $\endgroup$ Dec 2, 2014 at 15:32

1 Answer 1


Hoping I don't confuse you any further.

Variable cost is the cost that depends on how much you produce (hence, variable). So in your case that would be:

Now the average cost is the cost divided by how much is produced. The average variable cost then is:
$\text{AVC} = \frac{\text{VC}}{Q} =25 - 5Q + Q^2$

Now you want the minimum of the average cost. You find the minimum by deriving: $\frac{\partial \text{AVC}}{\partial Q} = 0; - 5 + 2Q = 0$
So you have that the minimum is at $Q=2.50$. Hurray. The AVC at the minimum is:
$\text{AVC}(2.5) = 25-5*(2.5) + (2.5)^2$, that is 18.75.

Now you also want the marginal cost, which is just:
$\text{MC}:\frac{\partial \text{TC}}{\partial Q} = 25 - 10Q + 3 Q^2$
$\text{MC}(2.5)=25-10*(2.5)+3*(2.5)^2$. And look at that we get back 18.75.

So apparently they are the same. Who would have thought.

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    $\begingroup$ I think it would be much better not to answer questions like this, as it will only attract more of them. $\endgroup$ Dec 2, 2014 at 15:56
  • $\begingroup$ To learn about Economics.SE's policy regarding answering homework questions, please refer to meta.economics.stackexchange.com/questions/24/… $\endgroup$ Dec 2, 2014 at 16:10
  • $\begingroup$ Also visit the link posted by @MartinVanderLinden to influence that policy! $\endgroup$
    – Ubiquitous
    Dec 2, 2014 at 17:08
  • $\begingroup$ sure, sure, let's keep these people out. We are already too many. $\endgroup$
    – CarrKnight
    Dec 3, 2014 at 3:15

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